# Fourier transform of derivative when integrating by parts

As seen in https://math.stackexchange.com/a/430885/634773, we can find the Fourier Transform of the derivative of a function through the anti-transform.

But if we do integration by parts, shouldn't it yield the same answer?

However, unlike what we see in https://math.stackexchange.com/a/430863/634773, the first term doesn't necessarily have a limit, or does it?

Sorry if I used bad english.

• Do you see how it works for $f(x) = e^{-x} 1_{x > 0}$ ? The derivative is $f'(x) = \delta(x)-e^{-x} 1_{x > 0}$ (a distribution, not a function), its primitive is $\int_a^x f'(y)dy = C + f(x)$, the Fourier transform of $\delta(x)$ is the boundary term you'll have when integrating by parts $\int_{-\infty}^\infty f(x) e^{- i \omega x}dx=\int_0^\infty e^{-x}e^{-i \omega x}dx$, and $\mathcal{F}[f'](\omega) = i \omega\mathcal{F}[f](\omega)$ – reuns Jun 23 '19 at 3:26
• Perfect English – gen-ℤ ready to perish Jun 23 '19 at 3:59
• There are many details being ignored in the posts you reference. My favorite justification was "why not?" – Disintegrating By Parts Jun 23 '19 at 17:15

One useful set of assumptions is that $$f,f'$$ are square integrable on $$\mathbb{R}$$ and $$f$$ is absolutely continuous. But then their transforms must be interpreted as limits in $$L^2$$ of the truncated transforms. This simple set of assumptions works because $$ff' \in L^1$$, which forces $$\int ff'dx = f^2/2$$ to have limits at $$\pm\infty$$, and those limits must be $$0$$ in order for $$f^2$$ to integral. Then you can integrate by parts to legitimately obtain $$\hat{f'}(\xi)=i\xi\hat{f}(\xi)$$.